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What is the nergy difference ( in kJ `mol^(-1)`) between the first and second shell fo H-atom if the first emission in the Lyman series occurs at ` lambda = 121.5` nm ? |
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Answer» I line of Lyman series is due to ` E_(2) - E_(1) = Delta E` ` therefore Delta E //mol e = ( h.c.N_(A))/(lambda)` `= (6.626 xx 10^(-34) xx 3 .0 xx 10^8 xx 6.023 xx 10^(23))/(121.5 xx 10^(-9))` `= 9.85 xx 10^(5) "J mol"^(-1) = 985k "J mol"^(-1)`. |
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